We are tasked with solving the logarithmic equation:

$\log_9(x) - \log_9(2x - 3) = \log_9(8)$

Step 1: Combine the logarithms

Using the logarithmic property $\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)$, the equation becomes:

$\log_9\left(\frac{x}{2x - 3}\right) = \log_9(8)$

Step 2: Remove the logarithms

Since the equation involves the same base ($\log_9$), we can exponentiate to eliminate the logarithms:

$\frac{x}{2x - 3} = 8$

Step 3: Solve the resulting equation

Multiply both sides by $(2x - 3)$ (assuming $2x - 3 \neq 0$):

$x = 8(2x - 3)$

Simplify:

$x = 16x - 24$

Rearrange terms to isolate $x$:

$24 = 15x \quad \Rightarrow \quad x = \frac{24}{15} = \frac{8}{5}$

Step 4: Check for extraneous solutions

For the logarithmic equation to be valid, the arguments of all logarithms must be positive:

• $x > 0$
• $2x - 3 > 0 \quad \Rightarrow \quad x > \frac{3}{2}$

The solution $x = \frac{8}{5} = 1.6$ satisfies $x > \frac{3}{2}$, so it is valid.

Final Answer:

The solution set is:

$\boxed{\left\{\frac{8}{5}\right\}}$

Let me know if you want further clarification or have additional questions.

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Related Questions:

1. What is the process for solving logarithmic equations in general?
2. How do you determine the domain of a logarithmic equation?
3. What are the common logarithmic properties used in equations like this?
4. What is the significance of checking for extraneous solutions in logarithmic equations?
5. How would the solution change if the base of the logarithms was different?

Tip:

Always check the domain of the logarithmic arguments before solving to ensure all solutions are valid.